Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint; as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g= 10m/s2)

When a 20 kg mass is attached to the midpoint of a horizontal rope, the rope sags, forming an angle θ with the horizontal. For the system to be in vertical equilibrium, the upward components of the tension (T) must balance the weight (W) of the mass. The weight is calculated as W = mg = 20 kg × 10 m/s² = 200 N [1]. The equilibrium equation is 2T sin(θ) = W, which simplifies to T = W / (2 sin(θ)). To completely straighten the rope, the angle θ must become 0 degrees. As θ approaches 0, sin(θ) also approaches 0. Mathematically, the tension T = 200 / (2 sin(0)) results in division by zero, meaning the tension required to eliminate the sag entirely is infinitely large. In practice, any finite tension will always result in some degree of sagging due to the vertical force of gravity acting on the mass.

Sources

1. [1] Science ,Class VIII . NCERT(Revised ed 2025) > Chapter 5: Exploring Forces > A step further > p. 75